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Math Help - Finding and evaluating integrals!!

  1. #1
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    Finding and evaluating integrals!!

    Hi Guys,
    I have attached the two questions i need help with....

    Really need a step by step explanation of how to complete please

    Please help!!!!
    Attached Thumbnails Attached Thumbnails Finding and evaluating integrals!!-image201303220002.jpg  
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Finding and evaluating integrals!!

    1. \int (2x-3)(x+4)dx.
    Expand the product and integrate term by term.
    2. \int_{-2}^{-1} \frac{dx}{(x-2)^3}
    Use the substitution t=x-2 (be careful with the integration bounds!)
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  3. #3
    Senior Member x3bnm's Avatar
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    Re: Finding and evaluating integrals!!

    For first question:

    \begin{align*}\int (2x-3)(x+4)\,\,dx =& \int (2x^2 + 5x - 12)\,\,dx\\ =& \frac{2x^3}{3} + \frac{5x^2}{2} - 12x + C\end{align*}

    You can see if the result is correct here http://www.wolframalpha.com/input/?i=integration+\int+(2x-3)(x%2B4)+dx

    For Second question let t = x - 2 \text{ and } \frac{dx}{dt} = 1:

    \text{for lower limit } x = -2 \text { So } t = -2 -2 = -4 \text{ for upper limit } x = -1 \text{ so } t = -1 -2 = -3


    \begin{align*}\int_{-2}^{-1} \frac{dx}{(x-2)^3} =& \int_{-4}^{-3} \frac{dt}{t^3}\\ =& \frac{t^{-2}}{-2}\Big]_{-4}^{-3}\\ =& \frac{(-3)^{-2}}{-2} - \frac{(-4)^{-2}}{-2}  \\ =& \frac{-1}{18} + \frac{1}{32} \\ =& \frac{-7}{288} \end{align*}

    You can check it here: http://www.wolframalpha.com/input/?i=integration+1%2F((x-2)^3)+dx+from+x+%3D+-2+to+-1
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